Question: $x^3+2y^2-xy=2$ Find the value of $\dfrac{dy}{dx}$ at the point $(0,-1)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac14$ (Choice B) B $-4$ (Choice C) C $4$ (Choice D) D $-\dfrac14$
We cannot isolate $y$ in order to define it as a function of $x$. Therefore, $x^3+2y^2-xy=2$ defines $y$ as a function of $x$ implicitly. To find $\dfrac{dy}{dx}$, we need to perform implicit differentiation. In implicit differentiation, we differentiate both sides of the equation according to $x$, and treat $y$ as an implicit function of $x$. [I need more explanation about implicit differentiation!] $\begin{aligned} x^3+2y^2-xy&=2 \\\\ \dfrac{d}{dx}(x^3+2y^2-xy)&=\dfrac{d}{dx}(2) \\\\ \dfrac{d}{dx}(x^3)+\dfrac{d}{dx}(2y^2)-\dfrac{d}{dx}(xy)&=0 \\\\ 3x^2+4y\cdot\dfrac{dy}{dx}-\left(1\cdot y+x\cdot\dfrac{dy}{dx}\right)&=0 \\\\ 3x^2+4y\cdot\dfrac{dy}{dx}-y-x\cdot\dfrac{dy}{dx}&=0 \end{aligned}$ Once we've completed the differentiation, we can arrange the equation so $\dfrac{dy}{dx}$ is isolated: $\begin{aligned} 3x^2+4y\cdot\dfrac{dy}{dx}-y-x\cdot\dfrac{dy}{dx}&=0 \\\\ \dfrac{dy}{dx}(4y-x)&=y-3x^2 \\\\ \dfrac{dy}{dx}&=\dfrac{y-3x^2}{4y-x} \end{aligned}$ Now we can plug the point $(0,-1)$ into the expression for $\dfrac{dy}{dx}$. $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{y-3x^2}{4y-x} \\\\ &=\dfrac{(-1)-3(0)^2}{4(-1)-0} \gray{x=0,\,\,y=-1} \\\\ &=\dfrac{-1}{-4} \\\\ &=\dfrac14 \end{aligned}$ In conclusion, the value of $\dfrac{dy}{dx}$ at the point $(0,-1)$ is $\dfrac14$.